(i) Suppose that \(\mu \)  is a (\(\beta +1\) )-Carleson measure and let \(d\nu (t)=(1-t)^{1-\beta }d\mu (t)\) . Using Lemma 3.2 of [1]}, we obtain that \(\nu \)  is a 2-Carleson measure. Then (REF ) and Proposition REF  imply that
\(\begin{split}\int _{\mathbb {D}} \overline{I_{\mu ,\alpha }(f)(r z)} g(r z) (1-|z|^2)^{\alpha -2}d A(z)=&\int _{[0,1)} \overline{f(t)} g(r^2t) d \mu (t)\\\le & C\Vert f\Vert _{{B}_{\beta }} \int _{[0,1)}|g(r^2t)| (1-t)^{1-\beta } d \mu (t) \\=&C\Vert f\Vert _{{B}_{\beta }} \int _{[0,1)}|g(r^2t)| d\nu (t)\\\le & C\Vert f\Vert _{{B}_{\beta }}\Vert g\Vert _{A^1},\quad f\in {B}_{\beta },~g\in A^1.\end{split}\) 

This and (REF ) show that \(\mathcal {I}_{\mu ,\alpha }\)  is a bounded operator from \({B}_{\beta }\)  into \({B}_{\alpha -1}\) .
On the other hand, suppose \(\mathcal {I}_{\mu ,\alpha }\)  is a bounded operator from \({B}_{\beta }\)  into \({B}_{\alpha -1}\) . For \(0<a<1\) , set
\(f_{a}(z)=\frac{1-a^2}{(1-az)^{\beta }}\) 
and
\(g_{a}(z)=\left(\frac{1-a^{2}}{(1-a z)^{2}}\right)^{2}.\) 
It is easy to check that there exists a positive constant \(C\)  which is dependent only on \(\beta \)  such that
\(\sup _{0<a<1}\Vert f_a\Vert _{{B}_{\beta }}\le C,\quad \sup _{0<a<1}\Vert g_a\Vert _{A^1}=1.\) 

Now the proof is similar to the proof of Theorem REF  (i).
(ii) For \(0<a<1\) , Let \(f_a\)  be defined as (i). Then \(\lbrace f_a\rbrace \)  is bounded sequence in \({B}_{\beta }\)  and \(\lim _{a\rightarrow 1}f_{a}(z)=0\)  on any compact subset of \(\mathbb {D}\) . From now on, the proof is analogous to the proof of Theorem REF  (ii) and we omit the details.
The operator \(\mathcal {H}_{\mu ,\alpha }\)  acting on Bloch type spaces
    In this section, we obtain the relationship between \(\mathcal {H}_{\mu ,\alpha }\)  and \(\mathcal {I}_{\mu ,\alpha }\) . We also obtain a necessary condition such that \(\mathcal {H}_{\mu ,\alpha }\)  is a bounded operator acting on Bloch type spaces for general cases. Then we characterize the measures \(\mu \)  for which the operator \(\mathcal {H}_{\mu ,\alpha }\)  (\(\alpha \ge 2\) ) is bounded (resp. compact) from Bloch type spaces into \({B}_{\alpha -1}\) . We begin with describing for which measures \(\mu \)  the operator \(\mathcal {H}_{\mu ,\alpha }\)  is well defined on the Bergman space \(A^p\)  (\(0<p<\infty \) ), which is a generalization of Theorem 2.1 in [2]}.
Proposition 3.1 
Suppose \(0<p<\infty \) , \(\alpha >0\)  and let \(\mu \)  be a positive Borel measure on \([0,1)\) . Then the power series in \((\ref {eq2})\)  defines an analytic function in \(\mathbb {D}\)  for every \(f\in A^p\)  in any of the three following cases.
\((i)\)
\(\mu \)  is a \(2/p\) -Carleson measure if \(0<p\le 1\) .
\((ii)\)
\(\mu \)  is a \(\left(\frac{2-(p-1)^2}{p}\right)\) -Carleson measure if \(1\le p\le 2\) .
\((iii)\)
\(\mu \)  is a \(1/p\) -Carleson measure if \(2\le p<\infty \) .

Furthermore, in such cases we have that
\(\mathcal {H}_{\mu ,\alpha }(f)(z)=\mathcal {I}_{\mu ,\alpha }(f)(z),\quad z\in \mathbb {D},~f\in A^p.\) 

Proof
(i) Suppose that \(\mu \)  is a \(2/p\) -Carleson measure. By Proposition 1 in [3]}, we obtain that there exists a positive constant \(C\)  which is independent of \(n,~k\)  such that
\(|\mu _{n,k,\alpha }|\le C\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}\frac{1}{(k+1)^{2/p}}.\) 
This and Theorem 4 in [4]} imply that, for every \(n\) ,
\(\sum _{k=0}^{\infty }\left|\mu _{n, k,\alpha }\right|\left|a_{k}\right| & \le C\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )} \sum _{k=0}^{\infty } \frac{\left|a_{k}\right|^{p}\left|a_{k}\right|^{1-p}}{(k+1)^{2/p}} \\& \le C\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )} \sum _{k=0}^{\infty } \frac{\left|a_{k}\right|^{p}\left|k+1\right|^{\frac{(2-p)(1-p)}{p}}}{(k+1)^{2/p}}\\&\le C\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )} \sum _{k=0}^{\infty } (k+1)^{p-3}\left|a_{k}\right|^{p}.\) 

This converges by Theorem 3 in [4]}. So the power series in (REF ) is well defined analytic function in \(\mathbb {D}\)  and
\(\sum _{k=0}^{\infty } \mu _{n, k,\alpha } a_{k}=\int _{[0,1)}\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}t^n f(t)d\mu (t).\) 
On the other hand, notice that for each \(z\in \mathbb {D}\) ,
\(&\sum _{n=0}^{\infty }(\int _{[0,1)}\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}t^n |f(t)|d\mu (t))|z|^n\\\le & \sum _{n=0}^{\infty }\left(\sum _{k=0}^{\infty } \mu _{n, k}|a_{k}|\right)|z|^n\le C \sum _{k=0}^{\infty }\left|a_{k}\right|^{p} (k+1)^{p-3}\sum _{n=0}^{\infty }\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}|z|^n\\\le &\frac{C}{(1-|z|)^{\alpha }}\sum _{k=0}^{\infty }\left|a_{k}\right|^{p} (k+1)^{p-3}<\infty .\) 

Therefore, we have
\(\mathcal {H}_{\mu ,\alpha }(f)(z) &=\sum _{n=0}^{\infty }\left(\int _{[0,1)}\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )} t^{n} f(t) d \mu (t)\right) z^{n} \\&=\int _{[0,1)} \sum _{n=0}^{\infty }\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}(t z)^{n} f(t) d \mu (t) \\&=\int _{[0,1)} \frac{f(t)}{(1-t z)^{\alpha }} d \mu (t)= \mathcal {I}_{\mu ,\alpha }(f)(z),\) 

for every \(z\in \mathbb {D}\)  and \(f\in A^p\) .
The proof of (ii) and (iii) is analogous to the proof of (i). We omit the details.
Notice that \({B}_{\beta }\subset \cap _{0<p<\infty }A^{p}\)  for \(0<\beta \le 1\) . We can easily obtain the following corollary by Proposition  REF
Corollary 3.1 
Suppose \(\alpha >0\) , \(0<\beta \le 1\) . If \(\mu \)  is a \(t\) -Carleson measure for some \(t>0\) , then for every \(f\in {B}_{\beta }\)  and \(z\in \mathbb {D}\) ,
\( \mathcal {H}_{\mu ,\alpha }(f)(z)=\mathcal {I}_{\mu ,\alpha }(f)(z).\) 

Proposition 3.2 
Suppose \(\alpha ,\beta >0\) , if \(\mu \)  is a \(\beta \) -Carleson measure, then for every \(f\in {B}_{\beta }\)  and \(z\in \mathbb {D}\) ,
\( \mathcal {H}_{\mu ,\alpha }(f)(z)=\mathcal {I}_{\mu ,\alpha }(f)(z).\) 

To proof the Proposition REF , we still need the following lemma, which can be found in [1]}.
Lemma 3.1 
Let \(f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}\in {B}_{\alpha }\)  for any \(\alpha >0\) . Then
\(\sup _{n}\sum _{k=2^{n}+1}^{2^{n+1}}\big |\frac{a_{k}}{k^{\alpha -1}}\big |^{2}<C\Vert f\Vert _{{B}_{\alpha }}^{2}.\) 

Proof of Proposition REF . Since \(\mu \)  is a \(\beta \) -Carleson measure, by Proposition 1 in [3]}, we have
\(|\mu _{n,k,\alpha }|\le C\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}\frac{1}{(k+1)^{\beta }}.\) 
Notice that \(\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}\sim n^{\alpha -1}\)  as \(n\rightarrow \infty \)  by Stirling's formula. It follows that there exist a constant \(C>0\)  such that
\(\sum _{n=0}^{\infty }\left(\sum _{k=2^{n}+1}^{2^{n+1}}\big |\frac{\mu _{n,k,\alpha }}{k^{1-\beta }}\big |^{2}\right)^{1/2}\le C\sum _{n=0}^{\infty }\left(n^{\alpha -1}\sum _{k=2^{n}+1}^{2^{n+1}}\frac{1}{k^2}\right)^{1/2}\le C\sum _{n=0}^{\infty }\frac{n^{(\alpha -1)/2}}{2^{n/2}}<\infty .\) 

This together with Lemma REF  yield that for every \(f\in {B}_{\beta }\) ,
\(\sum _{k=1}^{\infty }|\mu _{n,k,\alpha }a_k|&\le \sum _{k=1}^{\infty }\frac{\mu _{n,k,\alpha }}{k^{1-\beta }}\frac{|a_{k}|}{k^{\beta -1}}\\&\le \sum _{n=0}^{\infty }\left(\sum _{k=2^{n}+1}^{2^{n+1}}\big |\frac{\mu _{n,k,\alpha }}{k^{\beta -1}}\big |^{2}\right)^{1/2}\sup _{n}\left(\sum _{k=2^{n}+1}^{2^{n+1}}\big |\frac{a_{k}}{k^{\beta -1}}\big |^{2}\right)^{1/2}\\&<C\Vert f\Vert _{{B}_{\beta }}.\) 

From now on, arguing as in Proposition REF  (i), we can obtain the desired result.
Our next objective is to give a necessary condition such that the operator \(\mathcal {H}_{\mu ,\alpha }\)  is bounded on Bloch type spaces for general cases.
Theorem 3.1 
Suppose \(\alpha ,\beta ,\gamma >0\)  and \(\mathcal {H}_{\mu ,\alpha }\)  is a bounded operator from \({B}_{\beta }\)  into \({B}_{\gamma }\) .
(\(i\) )
If \(1<\beta <\infty \) , then the measure \(\mu \)  is an \((\alpha +\beta -\gamma -1/2)\) -Carleson measure if \(\alpha +\beta -\gamma -1/2>0\)  and \(\mu \)  is finite if \(\alpha +\beta -\gamma -1/2= 0\) .
(\(ii\) )
If \(0<\beta <1\) , then the measure \(\mu \)  is an \((\alpha -\gamma +1/2)\) -Carleson measure if \(\alpha -\gamma +1/2>0\)  and \(\mu \)  is finite if \(\alpha -\gamma +1/2= 0\) .


Proof
(i) Assume that the operator \(\mathcal {H}_{\mu ,\alpha }\)  is a bounded operator from \({B}_{\beta }\)  into \({B}_{\gamma }\) . For any \(0<\lambda <1\) , we set
\(f_{\lambda }(z)=\frac{1-\lambda ^{2}}{(1-\lambda z)^{\beta }}=\sum _{k=0}^{\infty }a_{k,\lambda }z^{k}\in {B}_{\beta }.\) 
Then
\(\mathcal {H}_{\mu ,\alpha }(f)(z)=\sum _{n=0}^{\infty }\big (\sum _{k=0}^{\infty }\mu _{n,k,\alpha }a_{k,\lambda }\big )z^n\in {B}_{\gamma }.\) 
It is easy to check that \(a_{k,\lambda }\sim ((1-\lambda ^2)k^{\beta -1}\lambda ^{k}).\) 
So Lemma REF  gives
\(\infty &>\sup _{j}\sum _{n=2^j+1}^{2^{j+1}}\left|\frac{\sum _{k=0}^{\infty }\mu _{n,k,\alpha }a_{k,\lambda }}{n^{\gamma -1}}\right|^2\\&\ge C\sup _{j}\sum _{n=2^j+1}^{2^{j+1}}\left|\frac{\sum _{k=0}^{\infty }(1-\lambda ^2)k^{\beta -1}\lambda ^{k}n^{\alpha -1}\int _{0}^{1}t^{n+k}d\mu (t)}{n^{\gamma -1}}\right|^2\\&\ge C\sup _{j}\sum _{n=2^j+1}^{2^{j+1}}\left|\frac{\sum _{k=0}^{\infty }(1-\lambda ^2)k^{\beta -1}\lambda ^{2k+n}n^{\alpha -1}\mu ([\lambda ,1))}{n^{\gamma -1}}\right|^2\\&\ge C\sup _{j}\sum _{n=2^j+1}^{2^{j+1}}\left|(1-\lambda ^2)\lambda ^{n}n^{\alpha -\gamma }\mu ([\lambda ,1))\sum _{k=0}^{\infty }k^{\beta -1}\lambda ^{2k}\right|^2\\&\ge C\sup _{j}\sum _{n=2^j+1}^{2^{j+1}}\left|\frac{(1-\lambda ^2)\lambda ^{n}n^{\alpha -\gamma }\mu ([\lambda ,1))}{(1-\lambda ^2)^{\beta }}\right|^{2}\\& \ge C\sup _{j}2^{j}\left|\frac{(1-\lambda ^2)\lambda ^{2^{j+1}}(2^{j})^{\alpha -\gamma }\mu ([\lambda ,1))}{(1-\lambda ^2)^{\beta }}\right|^{2}\) 

Choosing \(j\)  such that \(2^j\le \frac{1}{1-\lambda }<2^{j+1}\) , we have
\(\infty >\frac{C\mu ([\lambda ,1))}{(1-\lambda ^2)^{\alpha +\beta -\gamma -1/2}}.\) 
This is equivalent to saying that the measure \(\mu \)  is an \((\alpha +\beta -\gamma -1/2)\) -Carleson measure if \(\alpha +\beta -\gamma -1/2>0\)  and \(\mu \)  is finite if \(\alpha +\beta -\gamma -1/2\le 0\) . This completes the proof of Theorem REF .
(ii) The proof is similar to the proceeding one by taking \(f_{\lambda }=1\) .
Corollary 3.2 
Suppose \(\alpha ,\beta >0\)  and \(\mathcal {H}_{\mu ,\alpha }\)  is a bounded operator from \({B}_{\beta }\)  into \({B}_{\alpha -1}\) , then the measure \(\mu \)  is a \(3/2\) -Carleson measure for \(0<\beta <1\)  and \(\mu \)  is an \((\beta +1/2)\) -Carleson measure for \(1<\beta <\infty \) .

Now we can give the main results in this paper. By using Corollaries REF  and REF  together with Theorem REF  and Theorem REF  respectively, we can obtain the following two theorems.
Theorem 3.2 Suppose \(\alpha \ge 2\) , \(0<\beta <1\) . Let \(\mu \)  be a positive Borel measure on \([0,1)\) , then the following conditions are equivalent.
\((i)\)
\(\mathcal {H}_{\mu ,\alpha }\)  is a bounded operator from \({B}_{\beta }\)  into \({B}_{\alpha -1}\) .
\((ii)\)
\(\mathcal {H}_{\mu ,\alpha }\)  is a compact operator from \({B}_{\beta }\)  into \({B}_{\alpha -1}\) .
\((iii)\)
The measure \(\mu \)  is a 2-Carleson measure.


Theorem 3.3 Suppose \(\alpha \ge 2\) . Let \(\mu \)  be a positive Borel measure on \([0,1)\) , then
\((i)\)
\(\mathcal {H}_{\mu ,\alpha }\)  is a bounded operator from \({B}\)  into \({B}_{\alpha -1}\)  if and only if \(\mu \)  is a 1-logarithmic 2-Carleson measure.
\((ii)\)
\(\mathcal {H}_{\mu ,\alpha }\)  is a compact operator from \({B}\)  into \({B}_{\alpha -1}\)  if and only if \(\mu \)  is a vanishing 1-logarithmic 2-Carleson measure.


By applying Corollary REF , Proposition REF  and Theorem REF , we can obtain the following.
Theorem 3.4 Suppose that \(\alpha \ge 2\)  and \(1<\beta <\infty \) . Then
\((i)\)
\(\mathcal {H}_{\mu ,\alpha }\)  is a bounded operator from \({B}_{\beta }\)  into \({B}_{\alpha -1}\)  if and only if \(\mu \)  is a (\(\beta \) +1)-Carleson measure.
\((ii)\)
\(\mathcal {H}_{\mu ,\alpha }\)  is a compact operator from \({B}\)  into \({B}_{\alpha -1}\)  if and only if \(\mu \)  is a vanishing (\(\beta \) +1)-Carleson measure.


The final aim is to give a better necessary condition for some cases.
For \(0< p<\infty \) , a fuction \(f\in H(\mathbb {D})\)  is said to be of the class \(\mathcal {Q}_{p}\)  (cf. [8]}) in case
\(\sup _{w\in \mathbb {D}}\int _{\mathbb {D}}|f^{\prime }(z)|^{2}(-\log |\varphi _{w}(z)|)^{p}dA(z)<\infty ,\) 
where \(\varphi _{w}(z)=\frac{w-z}{1-\overline{w}z}\)  is a special M\(\ddot{\text{o}}\) bius map that interchanges the points 0 and \(w\) .
Lemma 3.2 \(\cite {Xiao-Q}\) 
Let \(0<p<\infty \)  and let \(f(z)=\sum _{k=0}^{\infty }a_kz^k\)  with \(a_k\)  nonnegative and nonincreasing. Then \(f\in \mathcal {Q}_{p}\)  if and only if \(\sup _{k}ka_{k}<\infty .\)

Theorem 3.5 Suppose \(0<\alpha \le 1\) , \(0<\beta <1\)  and \(0<p<\infty \) . If \(\mathcal {H}_{\mu ,\alpha }\)  is a bounded operator from \({B}_{\beta }\)  into \(\mathcal {Q}_{p}\) , then the measure \(\mu \)  is an \(\alpha \) -Carleson measure.

Proof
Assume that \(\mathcal {H}_{\mu ,\alpha }\)  is a bounded operator from \({B}_{\beta }\)  into \(\mathcal {Q}_{p}\) . We set \(f(z)=1\in {B_{\beta }}\) . Then
\( \mathcal {H}_{\mu ,\alpha }(f)(z)=\sum _{n=0}^{\infty }\big (\sum _{k=0}^{\infty }\mu _{n,k,\alpha }a_k\big )z^n=\sum _{n=0}^{\infty }\mu _{n,0,\alpha }z^n\in \mathcal {Q}_{p}.\) 
Notice that \(\mu _{n,0,\alpha }=\int _{[0,1)}\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}t^{n}d\mu (t)\)  (\(0<\alpha \le 1\) ) is positive and decreasing and \(\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}\sim n^{\alpha -1}\)  as \(n\rightarrow \infty \)  by Stirling's formula. For any \(0<\lambda <1\) , we take \(n\)  with \(1-\frac{1}{n+1}\le \lambda <1-\frac{1}{n}\) . By Lemma REF , we obtain that
\(\infty >n\mu _{n,0,\alpha }\ge Cn^{\alpha }\int _{\mathbb {D}}t^{n}d\mu (t)\ge Cn^{\alpha }\lambda ^{t}\int _{\lambda }^{1}d\mu (t)\ge \frac{C\mu ([\lambda ,1))}{(1-\lambda )^{\alpha }}.\) 
This shows that the measure \(\mu \)  is an \(\alpha \) -Carleson measure. The proof is completed.
It is well known that the class \(\mathcal {Q}_{p}\)  (\(p>1\) ) is equivalent to the Bloch space \({B}\)  and \(\mathcal {Q}_{1}\)  is just the space \(BMOA\)  (see [9]} for the more information of \(BMOA\) ). So we immediately have the following corollaries.
Corollary 3.3 Suppose \(0<\alpha \le 1\) , \(0<\beta <1\) . If \(\mathcal {H}_{\mu ,\alpha }\)  is a bounded operator from \({B}_{\beta }\)  into \({B}\) , then the measure \(\mu \)  is an \(\alpha \) -Carleson measure.

Corollary 3.4 Suppose \(0<\alpha \le 1\) , \(0<\beta <1\) . If \(\mathcal {H}_{\mu ,\alpha }\)  is a bounded operator from \({B}_{\beta }\)  into \(BMOA\) , then the measure \(\mu \)  is an \(\alpha \) -Carleson measure.

